given that A is an nxn matrix, how do you show that the characteristic polynomial of A is (-1)^n det(A)?

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- Nov 30th 2010, 12:56 PMalexandrabel90determinant
given that A is an nxn matrix, how do you show that the characteristic polynomial of A is (-1)^n det(A)?

- Nov 30th 2010, 02:54 PMtonio
Doing magic because the above is far from being true. In fact, $\displaystyle (-1)^n\det A$ is not even a polynomial of degree n, let alone

the characateristic polynomial of any nxn matrix. Read carefully again the question and repost.

Tonio

PS Perhaps there is something about some free coefficient or stuff...?(Giggle) - Nov 30th 2010, 04:47 PMalexandrabel90
the exact question is let A be an nxn matric and the trace of A is defined by tr(A)= a_11+a_22...+a_nn . Prove that the constant term of the characteristic polynomial is (-1)^n det(A)

- Nov 30th 2010, 05:12 PMDrexel28
- Nov 30th 2010, 05:27 PMalexandrabel90
yup, i know that..

so lets say, A is similar to an upper triangular matric so P(x) = (x-a_11)....(x-a_nn) = x^n + a_n-1 x^(n-1) +....+a_0 so P(0) = a_0 and then how do i continue? - Nov 30th 2010, 05:37 PMDrexel28
- Nov 30th 2010, 05:40 PMalexandrabel90
oh crap!!! im so blur! thanks